# Rubik’s Cube Not a Group?

although the $$3 \times 3 \times 3$$ is a great example of a mathematical group, larger cubes aren’t groups at all.

How can that be true? There is obviously an identity and it is closed, so that must mean that some moves aren’t invertible. But this seems unlikely to me.

• The moves on a "Rubik cube" of whatever size are a group action. May 13, 2012 at 0:35
• I don't know what the writer of that web page thought they meant, but the claim is incorrect. Of course it is a group.
– MJD
May 13, 2012 at 0:36
• The author seems to be implying that because you can have a "solved cube" that is not in its original position (by rotation of centers), then the cube is not "a group" because there are many "identities". But this is a confusion between a group and a group action, or else a confusion as to what constitutes an "identity"; or else you can view the cube as a suitable quotient. In short, the author is confused. May 13, 2012 at 0:45
• Don't worry too much about what you read online. A lot of it is wrong. Much of the rest is trivial. May 13, 2012 at 0:54
• @AndréNicolas On the other hand, the entirety of MSE and MathOverflow are online. Not saying your advice is good or bad, just an observation :-) May 13, 2012 at 1:02

The $$4\times 4\times 4$$ cube and higher aren't groups in the same sense that the $$3\times 3\times 3$$ cube is a group.

The set of reachable positions of a $$3\times 3\times 3$$ cube, viewed as functions from a $$54$$-element set (representing the locations of the stickers) to a $$6$$-element set (the colors of the stickers) form a group. The operation here is given by the following: For positions $$x, y$$ of a $$3\times 3\times 3$$ cube, let $$a_1a_2 \cdots a_n$$ be a sequence of moves which, starting from the identity, puts the cube into position $$x$$, and $$b_1b_2 \cdots b_m$$ the same for y. The product $$xy$$ is the state the cube is in after the sequence $$a_1 \cdots a_n b_1 \cdots b_m$$.

The fact that this operation indeed forms a group isn't as trivial as it first seems. The issue isn't with associativity, identity, or invertibility. Instead, it's with well-definedness. How do you know that the choice of sequences $$a_1 \cdots a_n$$ and $$b_1 \cdots b_m$$ doesn't make a difference?

For the $$3\times 3\times 3$$ cube, the way to solve this is viewing it as a subgroup of the symmetric group on the set of all $$54$$ stickers. This doesn't work for larger cubes, because it's possible to come up with moves that move some of the cubies around, without changing how the stickers show. The stickers can move, without their apparent colors changing. To see why this is impossible for a $$3\times 3\times 3$$ cube, note that any cubie is uniquely specified by its stickers, so any permutation of 2 stickers of the same color in the cube group must permute the cubies, which then cannot be the identity.

But on the $$4\times 4\times 4$$ cube, this fails. All $$4$$ of the starred stickers here, for example, are indistinguishable, in the sense that you could permute them and still have a solved state. I don't think a permutation of just these $$4$$ stickers is possible, but there are many permutations of indistinguishable cubies which can be done. According to Dustan Levenstein in the comments, any even permutation of these 4 stickers (or more generally of any center stickers) is possible.

The way to prove formally that the $$4\times 4\times 4$$ cube is not a group is to find a sequence of moves that acts as the identity on one configuration, but not on another configuration. This is pretty easy to do, but I don't remember the precise solution, so I'll omit it. (If anyone really wants such an example, comment and I can dig one up.)

It is true that if we labeled all of the stickers somehow, forming a so-called "supercube", and required that the labels also match up, then we would have a group. This group would be constructed in the same way as the $$3\times 3\times 3$$ group, as a subgroup of the symmetric group on the $$96$$ stickers of the $$4\times 4\times 4$$ cube.

This group acts on the set of positions of the $$4\times 4\times 4$$ cube transitively, but not freely. This, in group theoretical language, is why the $$4\times 4\times 4$$ cube positions do not form a group. We can still study the larger group, but we need to take into account that the action isn't as nice as in the $$3\times 3\times 3$$ case.

• Any even permutation of the middle stickers is possible, so in particular, an even permutation of just those those four stickers is possible. May 13, 2012 at 6:25
• What you are saying is that the action of the group of moves on the set of achievable positions of a cube with edges of length 4 or more has nontrivial stabilizers. Statements like "Rubik's Cube is a group" are so unclear that that it is pointless to discuss whether they are true or not! May 13, 2012 at 15:44
• What is true, I believe, is that in the $3\times3$ case, the stabilizer of an achievable position (which is the stabilizer of any achievable position) is a normal subgroup. So, if we want, we can define the Rubik's cube group for the $3\times3$ cube to be the quotient of the full group by this stabilizer. We can't do this for $4\times 4$ and higher since the stabilizers are not normal. May 13, 2012 at 16:37
• @DerekHolt Having nontrivial stabilizers is, of course, precisely what it means for an action to be non-free. As for the statement, of course it's unclear, that's why the OP wanted someone to interpret it. That doesn't mean it has no content. And this particular abuse of language is common among people who study twisty puzzles, so it's certainly worth knowing. May 13, 2012 at 17:20
• @WillOrrick You are correct, if you consider the orientation, and not just the color, of the stickers. I ignored this in my answer, but perhaps I should not have. The reason why I did is because the group of all moves which preserve the position, but not necessarily orientation, of each sticker is normal even for larger cubes, so we can mod out by it (and I did without explicitly saying so). For the 3x3x3 cube, this just so happens to be the stabilizer of every element, so the result is a normal action on the quotient. For the 4x4x4 cube, stabilizers are bigger. May 13, 2012 at 17:26

Arguments that the 3x3x3 is a group while the 4x4x4 isn't, based on the permutability of the center pieces in the latter case, are confused nonsense. By that standard, the 3x3x3 isn't a group either. Reason: although it is true that one cannot move the centers to other locations in the 3x3x3 case, one can rotate them in place. Consider this twist sequence: $(RLFR^-L^-F^2)^2$. If you use a cube with marked stickers, you will see that the net effect of this sequence is to rotate the $F$ center by 180 degrees. There is another sequence which rotates one center 90 degrees clockwise and another center 90 degrees counterclockwise. I see no reason to discount center rotations as 'movement'.

The real question is, when we say "the cube's group", what are we referring to? From all the articles I've read, the universal presumption is that the group is the set of all 'sticker' permutations (including permutations of the stickers' corners) achievable by any sequence of face/slice twists, regardless of visual distinguishability. In group-theoretic terms, there are three generally acknowledged constructs here:

• The free group generated by { $R, L, F, B, U, D$ }.

• The cube group which is the free group modulo the subgroup of all words which act as the identity on the sticker set - i.e the words which end up leaving all stickers exactly in their original position and orientation, not just in a perceptually indistinguishable state.

• The perceived state of the physical cube itself, which is the result of the action of the most recently applied element of the cube group.

I agree that the third construct isn't a group, but (to my knowledge) no one serious about cube theory ever thinks of that construct when referring to the cube as representing a group.